Finding the Expectation value for the ground state of a Hydrogen atom

John Paul Aseniero and Gibson T. Maglasang

For the particle in the state $\Psi$ , the expectation value of x is expressed as

$\langle x\rangle = \int_{-\infty}^{+\infty}x|\Psi(x,t)|^2dx$

where the expectation value is the average of repeated measurements on an ensemble of identically prepared systems.

In this article, we would like to find $\langle r\rangle$ , $\langle r^2\rangle$ , $\langle x\rangle$ and $\langle x^2\rangle$ for an electron in the ground state of a hydrogen atom and express it in Bohr radius.

a.) Calculating for the $\langle r\rangle$ and $\langle r^2\rangle$ ,

(i) Finding $\langle r\rangle$

The ground state wavefunction for the Hydrogen atom is given by

$\Psi_{100}(r\theta,\phi)=\frac{1}{\sqrt{\pi a^3}}e^{-r/a}$

Now getting the expectation value of r, we have

$\langle r\rangle = \int{r|\frac{1}{\sqrt{\pi a^3}}e^{-r/a}|^2}(r^2\sin\theta d\theta d\phi dr)$

$=\frac{1}{\pi a^3}\int_{0}^{\infty}\int_0^{2\pi}\int_0^{\pi}r^3 e^{-2r/a} \sin\theta d\theta d\phi dr$

$=\int_0^{\infty}r^3 e^{-2r/a}dr$

The above integration can now easily be facilitated by using the table of integral,

$\int_0^{\infty}r^3 e^{-2r/a}dr = 3!(\frac{a}{2})^4$ . (1)

Therefore,

$\langle r\rangle = \Big(\frac{4}{a^3}\Big)3!\Big(\frac{a}{2}\Big)^4$ .

(ii) For $\langle r^2\rangle$

Next is we find the value of $\langle r^2\rangle$ by using the same process employed in the previous exercise. We have,

$\langle r^2\rangle = \int r^2 |\frac{1}{\sqrt{\pi a^3}}e^{-r/a}|^2(r^2\sin\theta d\theta d\phi dr)$

$= \int_0^{\infty}\int_0^{2\pi}\int_0^{\pi}r^4\frac{1}{\pi a^3}e^{-2r/a}\sin\theta d\theta d\phi dr$

$=\frac{4}{a^3}\int_0^{\infty}r^4e^{-2r/a}dr$

Using again the table of integral used in (i) given in equation 1 to facilitate the integration, we get

$=\frac{4}{a^3}4!\Big(\frac{a}{5}\Big)^5$ .

Thus,

$\langle r^2\rangle=3a^2$ .

b) In the case of $\langle x\rangle$ and $\langle x^2\rangle$ , for electron in ground state of hydrogen atom, this requires no new integration since $r^2=x^2 + y^2 + z^2$ .

(i) For the calculation of $\langle x\rangle$

Now we have,

$\langle x\rangle = \int{x|\frac{1}{\sqrt{\pi a^3}}e^{-r/a}|^2}(r^2\sin\theta d\theta d\phi dr)$

but $x= r\sin\theta \cos\phi$ , it implies that

$\langle x\rangle=\frac{1}{\pi a^3}\int_{0}^{\infty}\int_0^{2\pi}\int_0^{\pi}(r\sin\theta\cos\phi)r^2 e^{-2r/a} \sin\theta d\theta d\phi dr$

Try to have a closer look at the integral of the $\phi$ part and evaluate it from 0 to $2\pi$ . Obviously we have,

$\int_0^{2\pi}\cos\phi d\phi=[sin\phi]|_0^{2\pi}=0$ .

Therefore,

$\langle x\rangle = 0$ .

(ii) However, for the $\langle x^2\rangle$

To find for $\langle x^2\rangle$ , we have the following calculation,

$\langle x^2\rangle = \int{x^2|\frac{1}{\sqrt{\pi a^3}}e^{-r/a}|^2}(r^2\sin\theta d\theta d\phi dr)$

$=\frac{1}{\pi a^3}\int_{0}^{\infty}\int_0^{2\pi}\int_0^{\pi}(r\sin\theta\cos\phi)^2r^2 e^{-2r/a} \sin\theta d\theta d\phi dr$

$= \frac{1}{\pi a^3}\int r^4e^{-2r/a}\sin^3\theta\cos^2\phi d\theta d\phi dr$

$= \frac{1}{\pi a^3}\int_0^{\infty}r^4e^{-2r/a}\int_0^{2\pi}\cos^2\phi d\phi\int_0^{2\pi}\sin^3\theta d\theta$

Note that the above integration is facilitated by the following integral formulas:

$\int_0^{2\pi} \sin^3\theta = \frac{4}{3}$ ,

$\int_0^{2\pi} \cos^2\phi d\phi = \pi$ .

Therefore,

$\langle x^2\rangle = \frac{1}{\pi a^3}\frac{4}{3}\pi \int_{0}^{\infty} r^4 e^{-2r/a}$

$= \frac{1}{3}\frac{4}{a^3}\int_0^{\infty} r^4 e^{-2r/a}dr$

$=\frac{1}{3}\langle r^2\rangle$ .

So,

$\langle x^2\rangle=a^2$ .

The expectation value for a ground state hydrogen atom are explicitly shown in this paper. The readers are also enjoined to calculate for the expectation value for momentum and see how they compare and contrast.

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Finding the Expectation value for the ground state of a Hydrogen atom

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